Question

Ajutor, va rog mult, rezolvarea sa fie pe foaie, clasa a 10 a, dau COROANA !​

Answer 1

$\it a)\ \sqrt{2x-1}=3 \Rightarrow (\sqrt{2x-1})^2=3^2 \Rightarrow 2x-1=9|_{+1} \Rightarrow 2x=10 \Rightarrow x=5\\ \\ \\ b)\ \sqrt[3]{\it2x-1}=3 \Rightarrow (\sqrt[3]{\it2x-1})^3=3^3 \Rightarrow 2x-1=27|_{+1} \Rightarrow 2x=28 \Rightarrow x=14\\ \\ \\ c)\ \sqrt{2x+3}=4x \Rightarrow (\sqrt{2x+3})=(4x)^2 \Rightarrow 2x+3=16x^2 \Rightarrow 16x^2-2x-3=0 \\ \\ a=16,\ \ b=-2,\ \ c=-3, \ \Delta' =1+48=49,\ \ \sqrt{\Delta'}=\sqrt{49}=7$

$\it x_{1,2}=\dfrac{1\pm7}{16};\ \ x_1=\dfrac{-6^{(2}}{16}=-\dfrac{3}{8};\ \ x_2=\dfrac{\ 8^{(8}}{16}=\dfrac{1}{2}$

$\it d)\ \sqrt[3]{x+2\sqrt2}=\sqrt2 \Rightarrow (\sqrt[3]{x+2\sqrt2})^3=(\sqrt2)^3 \Rightarrow \\ \\ \Rightarrow x+2\sqrt2=2\sqrt2|_{-2\sqrt2} \Rightarrow x=0$